Asked by Lindsay

A block weighing 70.7 N rests on a plane inclined at 21.6° to the horizontal. The coefficient of the static and kinetic frictions are 0.23 and 0.12 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping?

I STILL can't get this one. I've tried multiple answers, none of which are correct. Could someone please give me a little help?
Answered by drwls
When you are considering the force necessary to maintain a static condition, you use the static coefficient of friction. The maximum possible friction force is in that case
70.7*cos21.6 *0.23 = 15.12 N, The component of the gravity force down the plane is M g sin 23.6 = 70.7*0.400= 28.31 N.
Since the gravity component down the plane exceeds the maximum friction force by 13.19 N, that is the minimum force need to keep it from sliding.
Answered by Lindsay
Ok I understand that part.
Now it asks for the minimum magnitude of F that will start the block moving up the plane, and the magnitude of F that is required to move the block up the plane at constant velocity. Is it a similar process for both?
Answered by drwls
When you talk about a moving situation, the kinetic friction coefficient applies. The friction force is down the plane when trying to pull the block up the plane, so
70.7 sin 23.5 + 70.7 cos 23.5 * 0.12 = F
(For constant velocity, the forces balance)
Solve for F
Answered by Lindsay
Do you come up with an answer of 33.91? Because that's what I get.
Answered by drwls
No, I get 35.97 N
Answered by Lindsay
Why is it sin and cos of 23.5? I thought the angle was 21.6...
Answered by drwls
Yes. My mistake
Answered by Lindsay
Using 21.6 as the angle for both cos and sin, I come up with 33.91, which is the incorrect answer. Am I still doing something wrong?
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